Fanout:-Maximum Gate we connect to output of any Logic gate.

so there is limited number of gate we connect the output of Logic Gate.hence there is some minimum condition required for operation of Logic Gate(Transistor).calculation of Fanout depend on current.

so what's the minimum current required to Drive the Transistor(TTL Logic)?How we calculate?

Similarly for operation Of transistor minimum Voltage required 0.7V.


Got the answer. There is No limitation of input or base current, to turn on the transistor. even if the Base current is tending to zero but still transistor will be on. but the problem is, that the transistor will not be in saturation but in the active region which causes the output voltage to enter the confusing region. (related to the Noise margin of the TTL). is this the correct explanation?

  • \$\begingroup\$ What datasheet are you looking at that doesn't list input current? Most of them do. \$\endgroup\$
    – Hearth
    May 22 at 14:54
  • \$\begingroup\$ It's not only the current capability, it's the extra delay and possible extended linear region operation. You could drive (for example) 100 inputs in parallel, but the signal will be severely distorted. \$\endgroup\$ May 22 at 14:55
  • 1
    \$\begingroup\$ TTL specs look like this. But may be different depending upon the specific family chosen. \$\endgroup\$
    – jonk
    May 22 at 17:30
  • \$\begingroup\$ The problem with TTL is it needs much more current to turn the input off rather then on. Most will default to the input high. We used 1.6mA for each standard TTL input. \$\endgroup\$
    – Gil
    May 27 at 18:31
  • \$\begingroup\$ Standard TTL input, a little simplified, is a pull up resistor of 4.7k in series with a diode and a small capacitor around 10 pF to GND. An open TTL input is logic high, so no current if you apply high. \$\endgroup\$
    – Jens
    May 27 at 20:30


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